$\triangle ABC$ is an isosceles triangle inscribed in a circle with center $O$ and diameter $AD$ and $AB=AC$. The diameter intersects $BC$ at $E$, and $F$ is the midpoint of $OE$. Given that $BD\parallel FC$ and $BC=2 \sqrt[]{5}$, find the length of $CD$.

Here, $AD$ is perpendicular on $BC$
Hence, $BE=CE$
We can show that $$\triangle$$$BDE$ and $$\triangle$$$CFE$ are congruent.
So, $FE=ED=OF=\frac{1}{3}OD=\frac{1}{3}OB$
In the right angle triangle $OBE$,
$OE^2+BE^2=OB^2$
or,$4OF^2+5=9OF^2$
or,$OF=1=DE$
in rigth angle triangle $CED$,
$CE^2+DE^2=CD^2$
or,$CD^2=5+1$
or,$CD=\sqrt{6}$